Question

find the area of the region described by y≥2|x-3|+4 and x-2y≥-20

Answer 1

looks like some sort of triangle
http://www.wolframalpha.com/input/?i=+y%E2%89%A52 |x-3|%2B4+and+x-2y%E2%89%A5-20

Answer 2

x-3|%2B4+and+x-2y%E2%89%A5-20
what does this mean?

Answer 3

I think it was all one link

Answer 4

i think you start copying from the h to the last 0 there it is suppose to be a link if i'm not mistaken

Answer 5

Anyways we could try to draw this ourselves.

Answer 6

Let's look at the easiest part first.
How do you graph x-2y=-20?

Answer 7

y≥-x/(-2)-(-20)/(-2)

Answer 8

do you question marks in some diamond shape? I dont know if you see that, but i dont know how to make it stop

Answer 9

i have to keep refreshing the page for it to unshow

Answer 10

oh ok
is this right?
y≥-x/(-2)-(-20)/(-2)

Answer 11

\[x-2y \ge -20 \]
So I guess you subtracted x on both sides
\[-2y \ge -x-20\]
you should divide by the -2 on both sides
(when you divide both sides by a negative number remember to flip the inequality)
\[y \le \frac{x}{2}+10\]

Answer 12

well the direction of the inequality you have is not eating the right part
and you have to many negative over that on that 20/2 part

Answer 13

\[x-2y \ge -20 \\ -2y \ge -20-x \\ -2y \ge -x-20 \\ y \le \frac{-x}{-2}-\frac{20}{-2}\]

Answer 14

I think this is what you meant to write

Answer 15

oh ok i see my mistake

Answer 16

and you can cancel those negative's out in the x/2 term
and the same for that 20/2 part
\[y \le \frac{x}{2}+10 \]

Answer 17

ok

Answer 18

ok we have that where we can identify the slope and y-intercept of y=x/2+10 now

Answer 19

we could go ahead and look at the absolute value part

so we have
y=2|x-3|+4

if x-3>0 then |x-3|=x-3
if x-3<0 then |x-3|=-(x-3)

we are going to have to write y=2|x-3|+4 as a piecewise function

Answer 20

ok

Answer 21

or better yet if you know about translating graphs then we can skip that

Answer 22

do you know the vertex of y=2|x-3|+4?

Answer 23

y≥2|x|7 ??

Answer 24

?
the vertex of say y=a|x-h|+k is (h,k)
if a>0 then the absolute value opens up
if a<0 then the absolute value opens down

Answer 25

so what is the vertex of y=2|x-3|+4
try to use what I just said to answer

Answer 26

...i dont know...

Answer 27

sorry

Answer 28

ok so do you know that |x-3|=x-3 when x-3>0 and |x-3|=-(x-3) when x-3<0
because you know this then I guess we can go back to the piecewise thing

Answer 29

ok

Answer 30

so can you write y=2|x-3|+4 as a piecewise function?

Answer 31

y=2x-3+4 ?

Answer 32

remember |x-3|=x-3 if x-3>0
|x-3|=-(x-3) if x-3<0

Answer 33

do you know calculus? We can use that to find the area.
I was really hoping to draw the graph though but I don't think you can do that without knowing how to write absolute value function as a piecewise function or using the fact that the vertex is (3,4) and the absolute value function is open up because 2>0
I guess you can plug in values for x to see what y is (I hate this approach though because you have to plot enough points to get an almost accurate graph)

Answer 34

its ok i will just try to figure it our myself...thanks though :)